Patrick Stevens

There is a square arrangement made out of n elements on each side (n^2 elements total). You can put assign a value of +1 or -1 to any element. A function f is defined as the sum of the products of the elements of each row, over all rows and g is defined as the sum of the product of elements of each column, over all columns. Prove that, for n being an odd number, f(x)+g(x) can never be 0.

There is a very quick solution, similar in flavour to that famous dominoes puzzle. However, I didn’t come up with it immediately, and my investigation led down an interesting route.

Preliminary observations

It is easy to see that given a matrix of and , we have unchanged on reordering rows and columns, and on taking the transpose. This leads to a very useful lemma: are unchanged if we negate the corners of a rectangle in the matrix.

Specification of normal form

Given any four -1’s laid out at the corners of a rectangle, we may flip them all into 1’s without changing . Similarly, given any three -1’s on the corners of a rectangle, where the fourth corner is 1, we may flip to get a rectangle with one -1 and three 1’s.

Repeat this procedure until there are no rectangles with three or more corners -1. (Note that we might get a different answer depending on the order we do this in!) A Mathematica procedure to do this (expressed in a very disgusting way) is as follows.

Notice that columns which contain more than one -1 must not overlap, in the sense that no two columns with more than one -1 may have a -1 in the same row. Indeed, if they did, we’d have a submatrix somewhere of the form {{-1, -1}, {1, -1}}, which contradicts the “we’ve finished flipping” condition. Hence we may rearrange rows so that all -1’s appear together in contiguous columns.

We may then rearrange columns so that reading from the left, we see successive columns with decreasingly many -1’s. Rearrange rows again so that they appear stacked on top of each other.

We’ve ended up with a normal form: columns of -1’s, diagonally adjoined to each other, followed by rows of -1’s. (The following Mathematica code relies on the fact that SortBy is a stable sort.)

We haven’t shown that it’s unique yet, and indeed it’s not. As a counterexample, {{-1,1,1,1,1}, {-1,1,1,1,1}, {1,-1,1,1,1}, {1,-1,1,1,1}, {1,1,1,1,1}} is transformed into {{-1,1,1,1,1}, {-1,1,1,1,1}, {-1,1,1,1,1}, {-1,1,1,1,1},{1,1,1,1,1}} by a rectangle-flip.

This suggests a further improvement to the normal form: by flipping in this way, we may insist that any column of -1’s, other than the first, must contain only one -1. Indeed, if it contained two or more, we would flip two of them into the first column, rearrange so that all columns were contiguous -1’s again, and repeat.

What does our matrix look like now? It’s a column of -1, followed by some diagonal -1’s, followed by a row of -1. We’ll call this the canonical form, although I’ve still not shown uniqueness.

Restatement of problem

The problem then becomes: given a matrix in canonical form, show that cannot be 0.

Notice that if the long column is long, and there are diagonal -1’s, and the long row is long, and the matrix is , then , .

Hence .

Any choice of with yields a valid matrix. We therefore need to show that for all we have .

Solution

Reducing this mod 4, it is enough to show that . But we can easily case-bash the four cases which arise depending on the odd-even parity of , to see that in all four cases, the congruence does indeed not hold.

even: , but since is odd, this is not .

even, odd: , since is even so is a multiple of 4. isn’t even even, let alone divisible by .

odd: which is again not .

Summary

Once we had this canonical form, it was easy to find and therefore analyse the behaviour of . Next steps: prove that canonical forms are unique (perhaps using the fact that are invariant across forms, and showing a result along the lines that any two canonical forms with the same must be equivalent). I won’t do that now.